What is the integral from 0 to 1 of (1-8x^3+16x^7) ?

The integral from 0 to 1 of (1-8x^3+16x^7) is 1

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integral from 0 to 1 of (1-8x^3+16x^7)

\int_{0}^{1} (1 - 8 x^{3} + 16 x^{7}) d x

1

Solution steps

\int_{0}^{1} 1 - 8 x^{3} + 16 x^{7} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} 1 d x - \int_{0}^{1} 8 x^{3} d x + \int_{0}^{1} 16 x^{7} d x

\int_{0}^{1} 1 d x = 1

\int_{0}^{1} 8 x^{3} d x = 2

\int_{0}^{1} 16 x^{7} d x = 2

= 1 - 2 + 2

Simplify

= 1

\int_{1}^{2} (10 x^{3} + \frac{6}{x ^{2}}) d x

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\int_{4}^{6} x^{2} + 2 x - 8 d x

\int_{1}^{9} x^{\frac{1}{2}} d x

What is the integral from 0 to 1 of (1-8x^3+16x^7) ?
The integral from 0 to 1 of (1-8x^3+16x^7) is 1